An ideal Zener diode with breakdown voltage of \(3\,\text{V}\) is reverse biased with a negative input voltage \(V_1=-5\,\text{V}\).
The magnitude of voltage difference between points \(B\) and \(A\) is:
Show Hint
An ideal Zener diode maintains a constant voltage equal to its breakdown voltage.
Always check whether the applied reverse voltage exceeds the breakdown voltage.
If breakdown occurs, the voltage across the diode becomes constant.
Zener diodes are widely used as voltage regulators.
Step 1: Know the device. A Zener diode is built to operate steadily in reverse breakdown. Its breakdown voltage here is $V_Z = 3\,\text{V}$. Step 2: Read the applied voltage. The input is $V_1 = -5\,\text{V}$, a reverse bias of magnitude $5\,\text{V}$. Step 3: Check whether breakdown occurs. Since $|V_1| = 5\,\text{V} > V_Z = 3\,\text{V}$, the reverse voltage exceeds the breakdown level, so the Zener conducts in breakdown. Step 4: Apply the clamping property. Once in breakdown, an ideal Zener holds the voltage across itself fixed at $V_Z$, no matter how much larger the input is. Step 5: Voltage between B and A. The diode sits between $B$ and $A$, so the voltage difference across it is clamped to its breakdown value. \[ |V_{BA}| = V_Z = 3\,\text{V} \] Step 6: State the answer. The magnitude of the voltage difference between $B$ and $A$ is $3\,\text{V}$. \[ \boxed{3\,\text{V}} \]
